Question: The equation of a circle $C$ is $x^2+y^2-6x+2y-39 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-6x) + (y^2+2y) = 39$ $(x^2-6x+9) + (y^2+2y+1) = 39 + 9 + 1$ $(x-3)^{2} + (y+1)^{2} = 49 = 7^2$ Thus, $(h, k) = (3, -1)$ and $r = 7$.